Are all triangles isosceles 
I was reading a geometry book (HSM Coexeter's Geometry revisited) and I found something interesting about tritangent circles.
Let $r_a$ be the radius of excircle from $I_a$ 
then $ar(ABC)=rs=r_a(s-a)$....
$r=\text{inradius}$ and $s=\text{semiperimeter}$ and $a=BC$, $b=AC$, $c=AB$
Now,  As $\triangle AIY_b$ is similar to $\triangle AI_aY_a$
therefore 
$$\frac{AY_b}{AY_a}=\frac{s-c}{s}=\frac{r}{r_a}$$
which means $$rs=r_a(s-c)$$
Combining this with above equation gives 
$$a=c?$$
What's the flaw in the above reasoning?
NOTE:
$AZ_a=AY_a$ as these are tangents...
$AZ_a+AY_a=AB+CA+BX_a+CX_a=AB+BC+CA=2s$
$\therefore AZ_a=AY_a=s$
NOTE:
$AII_a,AY_bY_a$ are straight lines..join $I_aY_a$
Now $\angle AY_bI=\angle AY_aI_a=90$
and $\angle IAY_b=\angle I_aAY_a$
therefore triangles are similar.  
 A: So far as I can tell there is no flaw.  That diagram only exists for equilateral triangles. 
From Wolfram "Note that the three excircles are not necessarily tangent to the incircle, and so these four circles are not equivalent to the configuration of the Soddy circles".
In this diagram the three excircles are tangent to the incircle.  That can only occur if the triangle is equilateral.
For point $X_a$ to exist the incircle must be tangent to the excircle.  That can only happen if $AI_a$ is perpendicular to $BC$ and that can only happen if the triangle is isoceles.  This diagram of the excircles (tangent to all three extended lines of the triangle) being ALSO tangent to the incircle, can only occurs with equilateral triangles.
A: As others have noted, you seem to be inferring incidences from an unclear diagram. Here's an improved version:

Here, I hope you'll see that the points of tangency (which I've denoted $X$, $Y$, $Z$) of the incircle do not generally coincide with those ($X_a$, $Y_b$, $Z_c$) of the excircles. Moreover, neither of those points usually lies on the angle bisectors ($\overline{AI_a}$, $\overline{BI_b}$, $\overline{CI_c}$).
It is true, however, that when (and only when) two sides of the triangle are congruent, the points of tangency on the third side do coincide, and they lie on the corresponding angle bisector. So, your conclusion was "correct" based on your interpretation of the diagram; unfortunately (but somewhat understandably), your interpretation of the diagram was incorrect.
A: [Made an answer out of part of my comment.]
If you draw a triangle whose sides are very different from each other, you will see that the line between a vertex and the corresponding excircle center does not meet the point of tangency. In general, the points $A,X_a,I_a$ and similarly for $B$ and $C$ do not lie on a straight line.
